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I need to find the derivative of $\sin^2 (\sqrt{t})$ which I believe have done but the answer seems to be more simplified and I don't know how to arrive to it. Here are my steps

$$\begin{align} & \frac{d}{dt}(\sin \sqrt{t})^2 \\ & = 2(\sin \sqrt{t}) \cdot \frac{\cos \sqrt{t}}{2\sqrt{t}} \\ & = \frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}} \end{align}$$

The question is how does this simplify to $\frac{\sin 2\sqrt{t}}{2\sqrt{t}}$?

EDIT:

I forgot to add $2$ but I don't understand why top and bottom don't just cancel?

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3  
$\sin 2\theta=2\sin\theta\cos\theta$. So this will simplify to what you say, divided by $2$. –  1015 Apr 9 '13 at 13:21
    
@Julien but I don't have $2\sin\theta\cos\theta$? The $2$ on the top and $2$ on the bottom cancels out. –  gekkostate Apr 9 '13 at 13:22
    
$\frac{x}{y}=\frac{2x}{2y}$. Now you have the $2$ you miss at the numerator to apply the formula. Or if prefer, use $\sin\theta\cos\theta=\frac{\sin 2\theta}{2}$ directly. –  1015 Apr 9 '13 at 13:24
    
@RonGordon Sorry, I am a little confused. I plugged this into W|A and it has gives me the same simplified form that I arrived at. –  gekkostate Apr 9 '13 at 13:26
    
In response to the edit: It'd be sin to cancel them out. The expression is $\dfrac{\sin(2\sqrt t)}{2\sqrt t}$ rather than something else. –  Lord_Farin Apr 9 '13 at 13:34

3 Answers 3

up vote 2 down vote accepted

Hint: If $y=x$ in the formula $\sin \left(x+y\right)=\sin x \cos y + \cos x \sin y$ then $$ \sin(2x)= 2\cos(x)\sin(x). $$ Use this formula whit $x=\sqrt{t}$ to give \begin{align} \frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}}= & \frac{2\sin \sqrt{t} \cdot \cos \sqrt{t}}{2\sqrt{t}} \\ = & \frac{\sin (2\sqrt{t}) }{2\sqrt{t}} \end{align}

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Well for derivatives it's not necessary but for anti derivatives this one is really neat –  Dominic Michaelis Apr 9 '13 at 13:30

$$\frac{\sin \sqrt{t} \cdot \cos \sqrt{t}}{\sqrt{t}}=\frac{2\sin \sqrt{t} \cdot \cos \sqrt{t}}{2\sqrt{t}}=\frac{\sin (2\sqrt{t}) }{2\sqrt{t}}$$

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Your answer is correct and the target answer is missing a factor of $2$ in the denominator. Here is confirmation from Alpha

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